Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰa ⁴ᵗʰb ²ⁿᵈc ⟩, a > b > c
1st → a, 0th → b, a+b=1+6n
2nd → a, 0th → b, a+b=3
⟦0,1⟧ ∋ 2,4
⟨ ³ʳᵈ↓ ⟩ after ⟨→⟩
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ │ │ │ │ 1 │▒
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Step 2 │ │ │ │ │ 2 │ │ 1 │▒
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Step 3 │ │ │ │ │ 2 │ 6 │ 1 │▒
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Step 4 │ │ 0 │ │ │ 2 │ 6 │ 1 │▒
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Step 5 │ │ 0 │ 3 │ │ 2 │ 6 │ 1 │▒
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Step 6 │ │ 0 │ 3 │ 4 │ 2 │ 6 │ 1 │▒
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Step 7 │ 5 │ 0 │ 3 │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-03-26 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
By ✅「1st → a, 0th → b, a+b=1+6n」, we have
(1) [1st] + [0th] = 1 | 7.
If the above sum is 1, then { [1st], [0th] } = {0,1}, but then we cannot match ✅「⟦0,1⟧ ∋ 2,4」, which is a contradiction. It follows that
(2) [1st] + [0th] = 7.
On the other hand, by ✅「2nd → a, 0th → b, a+b=3」, we have
(3) { [2nd], [0th] } = {0,3} | {1,2}.
A fortiori, [0th] = 0|1|2|3. Observe that
(i) To satisfy (2), we need [0th] != 0;
(ii) ✅「⟦0,1⟧ ∋ 2,4」 implies that 2 is not in corners, so [0th] != 2;
(iii) If [0th] = 3, then (2) gives [1st] = 4, but then we cannot match ✅「⟦0,1⟧ ∋ 2,4」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
Therefore, we have [0th] = 1. Then, by ✅「2nd → a, 0th → b, a+b=3」 and (2), we get [2nd] = 2 and [1st] = 6 as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│ 0■│▒
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Step 1 │ │ │ │ │ │ │ 1 │▒
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Step 2 │ │ │ │ │ 2 │ │ 1 │▒
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Step 3 │ │ │ │ │ 2 │ 6 │ 1 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ 4 │ 5 │
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Next, we consider where to place 0. It is the minimum digit, so by ✅「⟨ ³ʳᵈ↓ ⟩ after ⟨→⟩」 it cannot be at 3rd. Similarly, by ✅「⟨ ⁶ᵗʰa ⁴ᵗʰb ²ⁿᵈc ⟩, a > b > c」, it cannot be at 6th or 4th. Consequently, 0 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 2 │ 6 │ 1 │▒
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Step 4 │ │ 0 │ │ │ 2 │ 6 │ 1 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to determine where to place 3. Note that the other idle digits (4 and 5) are both greater than it. Therefore, by ✅「⟨ ³ʳᵈ↓ ⟩ after ⟨→⟩」, we have 3 != [3rd], and by ✅「⟨ ⁶ᵗʰa ⁴ᵗʰb ²ⁿᵈc ⟩, a > b > c」, we have 3 != [6th]. It follows that 3 = [4th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
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│ │ 0 │ │ │ 2 │ 6 │ 1 │▒
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Step 5 │ │ 0 │ 3 │ │ 2 │ 6 │ 1 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, noting that there is only one way to match ✅「⟦0,1⟧ ∋ 2,4」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│ 3■│2nd│1st│0th│▒
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│ │ 0 │ 3 │ │ 2 │ 6 │ 1 │▒
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Step 6 │ │ 0 │ 3 │ 4 │ 2 │ 6 │ 1 │▒
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Step 7 │ 5 │ 0 │ 3 │ 4 │ 2 │ 6 │ 1 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.4